Average height of raindrops from height $h$ 
Raindrops drip from a spout at the edge of a roof and fall to the ground. Assume that the drops drip at a steady rate of $n$ drops per second (where $n$ is large) and that the height of the roof is $h$.
(c) What is the average height of these raindrops as $n$ grows larger when the first drop hits the ground?

First we need to find the time that the first raindrop hits the ground. We know from the projectile motion formula that
$$-\frac{1}{2}gt^2+h=0\Longrightarrow t=\sqrt{\frac{2h}{g}}$$
We then substitute it back into the first equation:
$$-\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2+h=0.$$
Then we solve for $h$:
$$h_{\text{avg}}=\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2$$
We add an additional factor since to find the limit of the average as n gets large. We get
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{2}g{\left(\frac{i}{n}\sqrt{\frac{2h}{g}}\right)}^2=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{i^2}{n^2}{h}.$$
Ultimately we solve
$$h\lim_{n\to\infty}\sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n},$$
which can be converted into
$$h\int_{0}^{1}x^2\,\mathrm dx=\frac{1}{3}h.$$
(This is the distance from the roof down. The real height is $\dfrac{2}{3}h$.)
Am I correct? If not, how do I find the average height?
 A: Let t be when the first (0th) drop hits the dirt.
At $n$ drops/sec, the next drop falls for $t - 1/n$ sec.
The $k$th drop falls for $t - {k/n}$ sec. and drops $d(k) = {g(t - k/n)}^{2}/{2}$ ft.
Its height $h(k) = h - d(k)$.
The average height is $\left(\sum\limits_{k=1}^{nt}h(k)\right)/nt$.  
A: You are getting the correct answer, despite making several errors.
Some errors cancel themselves out; in other cases you simply ignore the literal meaning of what you wrote and treat it as the kind of formula you should have written.
You are OK up to this point:
$$h_{avg}=\frac12 g{\left(\sqrt{\frac{2h}{g}}\right)}^2. \tag1$$
As you later seem to realize (because you use this fact to simplify things later),
$\frac12 g{\left(\sqrt{\frac{2h}{g}}\right)}^2 = h,$ and therefore Equation $(1)$ is simply saying that
$$ h_{avg}=h. $$
If you meant for $h_{avg}$ to be the average height of the drops above the ground,
we know that Equation $(1)$ is wrong; the average height is definitely less than $h.$
The equation before $h_{avg}$ is correct but is a waste of effort to write.
It simply says $-h + h = 0.$ It represents the fact that you correctly solved for $t$ in your earlier equation, $-\frac{1}{2}gt^2+h=0.$
Proceeding onward,
$$\frac{1}{n}\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{2}g{\left(\frac{i}{n}\sqrt{\frac{2h}{g}}\right)}^2. \tag2$$
The first obvious thing here is: why did you put the first $\frac1n$ outside the limit?  That makes no sense at all. What you've written says, first take the limit of something as $n$ gets large beyond bounds, and only then, after you have reduced the "limit" symbol and everything to the right of it to a single arithmetic expression with no $n$s in it (because that's what happens when you take a limit as $n$ goes to something), you're going to divide that expression by $n.$
Is that what you want to do?
Next, I assume that what you have in mind here is that there is a string of $n$ raindrops from the roof's edge to the ground, and that the $i$th drop in that string
(counting from the roof downward) has been falling for 
$\frac in\sqrt{\frac{2h}{g}}$ seconds.
This would be true if $\sqrt{\frac{2h}{g}} = 1,$ which is what we could conclude if we were given that there are $n$ raindrops between the roof and the ground, or if we were given that the roof was dripping $n$ raindrops per second and that it takes $1$ second for a drop to fall from the roof to the ground;
but that is not what we were given.
What we were actually given is that the roof drips $n$ drops per second and it is at a height $h$ above the ground.
The time for a drop to fall to the ground might be a lot less than one second, or a lot more.
In fact, you already showed that the time is $\sqrt{\frac{2h}{g}}.$
So in fact the $i$th drop (counting from the roof downward) has been falling for just $\frac in$ seconds, and the number of drops in the air at any given time is
$n \sqrt{\frac{2h}{g}}$ (rounded up or down to an integer, unless you allow "fractional raindrops").
To be fair, I think the wording of the problem is somewhat silly.
It could just as well have said something about the number of drops of water between the roof and the ground, and what happens if the rate increases so that there are a very large number of drops in the air.
If that number is $m,$ then $m \approx n \sqrt{\frac{2h}{g}}.$
It follows that the $i$th drop has been falling for
$$ \frac in \approx \frac im \sqrt{\frac{2h}{g}}$$
seconds, its distance from the roof is approximately
$$\frac12 g{\left(\frac im\sqrt{\frac{2h}{g}}\right)}^2,$$
and averaging this over all of the $m$ drops in the air, the average distance from the roof is
$$\frac 1m \sum_{i=1}^m \frac12 g{\left(\frac im\sqrt{\frac{2h}{g}}\right)}^2.$$
Now we want a limit as $n$ goes to $\infty,$
but since $m$ is (approximately) proportional to $n,$ that is the same as taking the limit as $m$ goes to $\infty,$
$$\lim_{m\to\infty}\frac 1m \sum_{i=1}^m \frac12 g{\left(\frac im\sqrt{\frac{2h}{g}}\right)}^2.\tag3$$
That's almost what you wrote, but using the total number of falling raindrops (not just one seconds' worth) and with the limit taken when it should be.
It seems you had the right intuition, just a little stumbling with the notation.
You applied correct (and useful) manipulations in the next couple of steps;
if we do the same thing to Equation $(3)$, we get
$$h\lim_{m\to\infty}\sum_{i=1}^m \frac{i^2}{m^2} \frac{1}{m},$$
and the fact that you wrote $n$ everywhere that $m$ occurs in this equation turns out to have no effect on the final result;  the $\lim$ notation creates what is called a "bound variable," and technically regardless how the name of that variable is used outside the limit, inside the limit it is a different variable.
There's a little stumbling again here:
$$ h \int_0^1 n^2.$$
You did not indicate the variable over which you intend to integrate.
I suppose you meant to integrate over $n,$ like this: 
$\int_0^1 n^2 \,dn.$
But $n$ was introduced earlier as the number of raindrops falling per second,
which is a large number, not between $0$ and $1.$
Someone might argue that $n$ is supposed to be a constant within this integral, that is, that the integral should be read as 
$\int_0^1 n^2 \,dx = n^2.$
The way to avoid this kind of objection is not to use $n$ in this way. 
What you could say is that if you set $x_i = \frac in$ in the sum, then the sum approximates the integral
$$\int_0^1 x^2 \,dx = \frac 13.$$
Again, the intuition was correct. You just might want to take a little more care how you write the equations, to make sure people know what you meant and agree that the math is correct.
A: $\def\peq{\mathrel{\phantom{=}}{}}$Assume instead that there are $n$ drops per $T$ seconds so that $n$ becomes dimensionless, thus the time interval between two drops is $Δt = \dfrac{T}{n}$.
It takes $t_1 = \sqrt{\dfrac{2h}{g}}$ for the first drop to hit the ground, and when it hits, there are $N = \left[ \dfrac{t_1}{Δt} \right] + 1$ drops in the air (including the first drop) and the $k$-th drop has traveled $t_k = t_1 - (k - 1)Δt$, which implies its current height is $h - \dfrac{1}{2} g t_k^2$. Therefore, the average height is\begin{align*}
&\peq \frac{1}{N} \sum_{k = 1}^N \left(h - \frac{1}{2} g t_k^2 \right) = h - \frac{g}{2N} \sum_{k = 1}^N (t_1 - (k - 1)Δt)^2\\
&= h - \frac{g}{2N} \left( N t_1^2 - 2t_1 Δt \sum_{k = 1}^N (k - 1) + (Δt)^2 \sum_{k = 1}^N (k - 1)^2 \right)\\
&= h - \frac{g}{2N} \left( N t_1^2 - N(N - 1) t_1 Δt + \frac{1}{6} (2N - 1)N(N - 1) (Δt)^2 \right)\\
&= h - \frac{1}{2} g t_1^2 + \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2\\
&= \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2.
\end{align*}
Note that $N Δt → t_1$ as $n → ∞$, thus\begin{align*}
&\peq \lim_{n → ∞} \left( \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2 \right)\\
&= \frac{1}{2} g t_1^2 - \frac{1}{6} g t_1^2 = \frac{2}{3} h.
\end{align*}
